Difference between revisions of "Proof that 2=1"

Revision as of 13:43, 27 January 2022

Proof

1) $a = b$ . Given.

2) $a^2 = ab$ . Multiply both sides by a.

3) $a^2-b^2 = ab-b^2$ . Subtract $b^2$ from both sides.

4) $(a+b)(a-b) = b(a-b)$ . Factor both sides.

5) $(a+b) = b$ . Divide both sides by $(a-b)$

6) $a+a = a$ . Substitute $a$ for $b$ .

7) $2a = a$ . Addition.

8) $2 = 1$ . Divide both sides by $a$ .

Error

Usually, if a proof proves a statement that is clearly false, the proof has probably divided by zero in some way.

In this case, the quantity of $a-b$ is $0$ as $a = b$ , since one cannot divide by zero, the proof is incorrect from that point on.

Thus, this proof is false.

Note:

If this proof were somehow true all of mathematics would collapse. Simple arithmetic would yield infinite answers. This is why one cannot divide by zero.

Alternate Proof

Consider the continued fraction $3-\frac{2}{3-\frac{2}{3-\frac{2}{3- \cdots}}}.$ If you set this equal to a number $x$ , note that $3-\frac{2}{x}=x$ due to the fact that the fraction is infinitely continued. But this equation for $x$ has two solutions, $x=1$ and $x=2.$ Since both $2$ and $1$ are equal to the same continued fraction, we have proved that $2=1.$ QED.

@@ Line 28: / Line 28: @@
 ==Alternate Proof==
-Consider the continued fraction <math>3-\frac{2}{3-\frac{2}{3-frac{2}{3- \cdots}}}.</math>
+Consider the continued fraction <math>3-\frac{2}{3-\frac{2}{3-\frac{2}{3- \cdots}}}.</math> If you set this equal to a number <math>x</math>, note that
+<math>3-\frac{2}{x}=x</math> due to the fact that the fraction is infinitely continued. But this equation for <math>x</math> has two solutions, <math>x=1</math> and <math>x=2.</math> Since both <math>2</math> and <math>1</math> are equal to the same continued fraction, we have proved that <math>2=1.</math> QED.

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